What is (\partial)/(\partial y)(xcos(y)+ye^x) ?

The answer to (\partial)/(\partial y)(xcos(y)+ye^x) is -xsin(y)+e^x

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(\partial)/(\partial y)(xcos(y)+ye^x)

\frac{\partial}{\partial y} (x cos (y) + y e^{x})

- x sin (y) + e^{x}

Solution steps

\frac{\partial}{\partial y} (x cos (y) + y e^{x})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (x cos (y)) + \frac{\partial}{\partial y} (y e^{x})

\frac{\partial}{\partial y} (x cos (y)) = - x sin (y)

\frac{\partial}{\partial y} (y e^{x}) = e^{x}

= - x sin (y) + e^{x}

\frac{d}{d x} (\frac{x ^{2} - 1}{2 x})

x \to - 2 lim (\frac{2 - x ^{2}}{4 x})

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0, y (0) = 6

\frac{1}{3 ( 4 - x ^{3} ) ^{4}}

\int \frac{3 x + 1}{x ^{2}} d x

What is (\partial)/(\partial y)(xcos(y)+ye^x) ?
The answer to (\partial)/(\partial y)(xcos(y)+ye^x) is -xsin(y)+e^x